I'm not able to solve the above problem. Please have a look at my attempt at the solution, where I'm stuck - also, your methods to solve the problem are welcome and you're requested to post them in the answers section.
For starters, here's what I did -
I constructed the incircle of ∆ABC, and named its intersection points with the circle as A', B' and C'. Clearly, AC' = AB'. Also, I noticed that the centre of the bigger circle, I, is collinear with D and G as the angle subtended at A is 90°. The inradius can be found easily (it's 15), by using Pythagoras Theorem to find the third side and using the well known result ∆ = rs, where ∆ is the triangle's area, r is its inradius and s is its semiperimeter.
Now, how should I proceed? Could someone please point me in the right direction, or provide a solution to the problem (which would probably help me understand the problem better)?
Thanks a lot!
$$a=\sqrt{55^2+48^2}=73.$$ Thus, $$r=\frac{55+48-73}{2}=15$$ and since $$\measuredangle BAI=45^{\circ},$$ we obtain $$AD=2r=30.$$ Similarly, $$AG=EF=30,$$ which gives that $DAFE$ is trapezoid and since $$\measuredangle DEG=\measuredangle DAG=90^{\circ},$$ we obtain $EG\perp AF$.
Similarly, $DF\perp AE$, which says $AH\perp EF$ and since $\measuredangle EAF=45^{\circ},$ we obtain: $$AH=EF\cot\measuredangle EAF=30\cot45^{\circ}=30.$$ By the way, there is a proof that $AH=EF$ without trigonometry.